Tìm giới hạn sau:
a) \(lim\frac{6n - 1}{3n +2}\);
b) \(lim\frac{3n^{2}+n-5}{2n^{2}+1}\);
c) \(lim\frac{3^{n}+5.4^{n}}{4^{n}+2^{n}}\);
d) \(lim\frac{\sqrt{9n^{2}-n+1}}{4n -2}\).
Câu a:
\(lim\frac{6n - 1}{3n +2}= lim\frac{6 - \frac{1}{n}}{3 +\frac{2}{n}}=\frac{6}{n}=2\)
Câu b:
\(lim\frac{3n^{2}+n-5}{2n^{2}+1} =lim\frac{3 +\frac{1}{n}-\frac{5}{n^{2}}}{2+\frac{1}{n^{2}}}=\frac{3}{2}\)
Câu c:
\(lim\frac{{{3^n} + {{5.4}^n}}}{{{4^n} + {2^n}}} = lim\frac{{{{\left( {\frac{3}{4}} \right)}^n} + 5}}{{1 + {{\left( {\frac{1}{2}} \right)}^n}}} = \frac{5}{1} = 5\)
Câu d:
\(lim\frac{{\sqrt {9{n^2} - n + 1} }}{{4n - 2}} = lim\frac{{n\sqrt {9 - \frac{1}{n} + \frac{1}{{{n^2}}}} }}{{n(4 - \frac{2}{n})}} = lim\frac{{\sqrt {9 - \frac{1}{n} + \frac{1}{{{n^2}}}} }}{{4 - \frac{2}{n}}} = \frac{{\sqrt 9 }}{4} = \frac{3}{4}\)
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